Question

Rhino viruses typically cause common colds. In a test of the effectiveness of​ echinacea, 45 of the 52 subjects treated with echinacea developed rhinovirus infections. In a placebo​ group, 81 of the 96 subjects developed rhinovirus infections. Use a 0.05 significance level to test the claim that echinacea has an effect on rhinovirus infections.

Answer 1

Answer)

Null hypothesis Ho : p1 = p2

Alternate hypothesis Ha : p1 not equal to p2

N1 = 52, P1 = 45/52

N2 = 96, P2 = 81/96

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 45

N1*(1-p1) = 7

N2*p2 = 81

N2*(1-p2) = 15

All the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = 0.35

From z table, P(Z>0.35) = 0.3632

As the test is two tailed

So, P-Value = 2*0.3632 = 0.7264

As the obtained P-Value is greater than the given significance 0.05

We fail to reject the null hypothesis Ho

We do not have enough evidence to conclude that echinacea has an effect on rhinovirus infections.